Chemistry - Chemical Kinetics Question with Solution | TestHub

Let the order of reaction with respect to $\mathrm{A}$ is $x$ and with respect to $B$ is $y$. Thus, $$\text{ rate }=\mathrm{k}[\mathrm{A}{]}^{\mathrm{x}}[\mathrm{B}{]}^{\mathrm{y}}$$ For the given cases, (I) rate $=\mathrm{k}(0.1{)}^{\mathrm{x}}(0.1{)}^{\mathrm{y}}=6.0\times 10^{-3}$ (II) rate $=\mathrm{k}(0.3{)}^{\mathrm{x}}(0.2{)}^{\mathrm{y}}=7.2\times 10^{-2}$ (III) rate $$=\mathrm{k}(0.3{)}^{\mathrm{x}}(0.40{)}^{\mathrm{y}}=2.88\times 10^{-1}$$ (IV) rate $=\mathrm{k}(0.4{)}^{\mathrm{x}}(0.1{)}^{\mathrm{y}}=2.40\times 10^{-2}$ On dividing Eq. (I) by (IV), we get $${\left(\frac{0.1}{0.4}\right)}^x{\left(\frac{0.1}{0.1}\right)}^y=\frac{6.0\times 10^{-3}}{2.4\times 10^{-2}}$$ or $${\left(\frac{1}{4}\right)}^x={\left(\frac{1}{4}\right)}^1$$ $$\therefore \mathrm{x}=1$$ On dividing Eq. (II) by (III), we get $$\begin{array}{rlrl}& & {\left(\frac{0.3}{0.3}\right)}^x{\left(\frac{0.2}{0.4}\right)}^y& =\frac{7.2\times 10^{-2}}{2.88\times 10^{-1}}\\ \text{ or }& & {\left(\frac{1}{2}\right)}^y& =\frac{1}{4}\\ \text{ or }& & {\left(\frac{1}{2}\right)}^y& ={\left(\frac{1}{2}\right)}^2\\ \therefore & & y& =2\end{array}$$ $\therefore$ Thus, rate law is, $$\text{ rate }=\mathrm{k}[\mathrm{A}{]}^1[\mathrm{B}{]}^2\text{ or }=\mathrm{k}[\mathrm{A}][\mathrm{B}{]}^2$$